PHYS 1443 – Section 003 Lecture #17AnnouncementsConditions for EquilibriumMore on Conditions for EquilibriumCenter of Gravity RevisitedExample 12.1Example 12.1 ContinuedExample 12.2Example 12.3Example 12.4How did we solve equilibrium problems?Wednesday, Nov. 13, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #17Wednesday, Nov. 13, 2002Dr. Jaehoon Yu1. Conditions for Mechanical Equilibrium2. Center of Gravity in Mechanical Equilibrium 3. Elastic Properties of Solids•Young’s Modulus•Shear Modulus•Bulk ModulusToday’s homework is homework #17 due 12:00pm, Wednesday, Nov. 20!!Wednesday, Nov. 13, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu2Announcements•Quiz–Quiz problem number 3 was incorrectly formulated everyone gets credit for this problem–Average: 70.3–Many of you had 100 points–Volunteer to solve quiz problems for us? •Congratulations, Matt Andrews, for winning 7th place in International Computer Programming ContestWednesday, Nov. 13, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu3Conditions for EquilibriumWhat do you think does the term “An object is at its equilibrium” mean?0FThe object is either at rest (Static Equilibrium) or its center of mass is moving with a constant velocity (Dynamic Equilibrium). Is this it? When do you think an object is at its equilibrium?Translational Equilibrium: Equilibrium in linear motion The above condition is sufficient for a point-like particle to be at its static equilibrium. However for object with size this is not sufficient. One more condition is needed. What is it? Let’s consider two forces equal magnitude but opposite direction acting on a rigid object as shown in the figure. What do you think will happen?CMddF-FThe object will rotate about the CM. The net torque acting on the object about any axis must be 0. For an object to be at its static equilibrium, the object should not have linear or angular speed. 00CMv0Wednesday, Nov. 13, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu4More on Conditions for EquilibriumTo simplify the problems, we will only deal with forces acting on x-y plane, giving torque only along z-axis. What do you think the conditions for equilibrium be in this case? The six possible equations from the two vector equations turns to three equations.What happens if there are many forces exerting on the object?Net torque about O0F00xF0zOF1F4F3F2F5r5O’r’If an object is at its translational static equilibrium, and if the net torque acting on the object is 0 about one axis, the net torque must be 0 about any arbitrary axis. 0321FFFFONet Force exerting on the object''rrriiPosition of force Fi about O’Net torque about O’'O'O0yF332211FrFrFriiFr0 2211'' FrrFrr iiiFrFr '2211'' FrFr 0'rFriiO0Wednesday, Nov. 13, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu5Center of Gravity RevisitedWhen is the center of gravity of a rigid body the same as the center of mass?CMxUnder the uniform gravitational field throughout the body of the object.CMm1g1CoGm2g2m3g3migiLet’s consider an arbitrary shaped objectThe center of mass of this object isLet’s now examine the case with gravitational acceleration on each point is giSince the CoG is the point as if all the gravitational force is exerted on, the torque due to this force becomes CoGxgmgm 2211If g is uniform throughout the bodyGeneralized expression for different g throughout the bodyiiimxmMxmiiCMyiiimymMymii222111xgmxgm CoGgxmm 21 gxmxm 2211CoGxiiimxmCMxWednesday, Nov. 13, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu6Example 12.1A uniform 40.0 N board supports a father and daughter weighing 800 N and 350 N, respectively. If the support (or fulcrum) is under the center of gravity of the board and the father is 1.00 m from CoG, what is the magnitude of normal force n exerted on the board by the support?Since there is no linear motion, this system is in its translational equilibriumFDn1m xTherefore the magnitude of the normal force nDetermine where the child should sit to balance the system.The net torque about the fulcrum by the three forces are Therefore to balance the system the daughter must sitxxF0yFgMB0gMFgMDnmgMgMDF00.1mm 29.200.135 08000 gMB00.1 gMFxgMD0N119035 080 00.40 MBgMFgMFgWednesday, Nov. 13, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu7Example 12.1 ContinuedDetermine the position of the child to balance the system for different position of axis of rotation.Since the normal force is The net torque about the axis of rotation by all the forces are ThereforexnThe net torque can be rewritten What do we learn?No matter where the rotation axis is, net effect of the torque is identical.FDnMBgMFg MFg1m xx/2Rotational axis2/xgMB0gMgMgMDFB 2/00.1 xgMF2/xn 2/xgMD2/xgMB 2/00.1 xgMF 2/xgMgMgMDFB2/xgMDxgMgMDF 00.10mgMgMDF00.1mm 29.200.135080 0Wednesday, Nov. 13, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu8Example 12.2A person holds a 50.0N sphere in his hand. The forearm is horizontal. The biceps muscle is attached 3.00 cm from the joint, and the sphere is 35.0cm from the joint. Find the upward force exerted by the biceps on the forearm and the downward force exerted by the upper arm on the forearm and acting at the joint. Neglect the weight of forearm.xFSince the system is in equilibrium, from the translational equilibrium conditionFrom the rotational equilibrium conditionOFBFUmgdlThus, the force exerted by the biceps muscle isdFBForce exerted by the upper arm isUF0yFmgFFUB0lmgdFFBU 00lmg BFdlmg N58300.30.350.50mgFBN5 3 30.50583 Wednesday, Nov. 13, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu9Example 12.3A uniform horizontal beam with a length of 8.00m and a weight of 200N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of 53.0o with the horizontal. If 600N person stands 2.00m from the
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